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\(|12,1.x+12,1.0,1|=|12,1.\left(x+0,1\right)|\)
=> 12,1.(x+0,1)=21,1 hoặc -21,1
=> x+0,1=\(\frac{211}{121}\)hoặc\(\frac{-211}{121}\)
=. x = \(\frac{1989}{1210}\)hoặc \(-1,844\)
\(a.\)
\(\left|0,2x-3,1\right|=6,3\)
\(\Rightarrow\left[\begin{array}{nghiempt}0,2x-3,1=6,3\\0,2x-3,1=-6,3\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}0,2x=6,3+3,1\\0,2=-6,3+3,1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}0,2x=9,4\\0,2x=-3,2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=9,4:0,2\\x=-3,2:0,2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=47\\x=-16\end{array}\right.\)
Vậy : \(x\in\left\{-16;47\right\}\)
\(b.\)
\(\left|12,1x+12,1.0,1\right|=12,1\)
\(\Leftrightarrow\left|12,1.\left(x+0,1\right)\right|=12,1\)
\(\Rightarrow\left[\begin{array}{nghiempt}12,1\left(x+0,1\right)=12,1\\12,1\left(x+0,1\right)=-12,1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+0,1=12,1:12,1\\x+0,1=-12,1:12,1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+0,1=1\\x+0,1=-1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=1-0,1\\x=-1-0,1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0,9\\x=-1,1\end{array}\right.\)
Vậy : \(x\in\left\{-1,1;-,9\right\}\)
\(c.\)
\(\left|0,2x-3,1\right|+\left|0,2x=3,1\right|=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}0,2x-3,1=0\\0,2+3,1=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}0,2x=0+3,1\\0,2x=0-3,1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}0,2x=3,1\\0,2x=-3,1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=3,1:0,2\\x=-3,1:0,2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=15,5\\x=-15,5\end{array}\right.\)
Vậy : \(x\in\left\{-15,5;15,5\right\}\)
a ) \(\left|0,2.x-3,1\right|=6,3\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}0,2x-3,1=6,3\\0,2x-3,1=-6,3\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}0,2x=9,4\\0,2x=-3,2\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=47\\x=-16\end{array}\right.\)
Vậy ........
b ) \(\left|12,1.x+12,1.0,1\right|=12,1\)
\(\Leftrightarrow\left|12,1.\left(x+0,1\right)\right|=12,1\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}12,1.\left(x+0,1\right)=12,1\\12,1.\left(x+0,1\right)=-12,1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x+0,1\right)=1\\\left(x+0,1\right)=-1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0,9\\x=-1,1\end{array}\right.\)
Vậy ...........................
c ) \(\left|0,2.x-3,1\right|+\left|0,2.x+3,1\right|=0\)
=> \(\left[\begin{array}{nghiempt}0,2x-3,1=0\\0,2x+3,1=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}0,2x=3,1\\0,2x=-3,1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=15,5\\x=-15,5\end{array}\right.\)
Vậy .............................
Có hai trường hợp:
Nếu phần trị tuyệt đối bằng -12,1 thì x = -1,1
Nếu phần trị tuyệt đối bằng 12,1 thì x= 0,9
1) Tính nhanh
a) \(6,5+1,2+3,5-5,2+6,5-4,8\)
\(=\left(6,5+3,5\right)-\left(5,2+4,8\right)+\left(1,2+6,5\right)\)
\(=10-10+7,7\)
\(=7,7\)
x + x : 0,2 = 1,35
x * 1 + x * 5 = 1,35
x * ( 1 + 5 ) = 1,35
x * 6 = 1,35
x = 1,35 : 6
x = 0,225
hok tốt nha ^_^
a) |0,2x - 3,1| = 6,3
\(0,2x-3,1=\pm6,3\)
Th1:
0,2x - 3,1 = 6,3
0,2x = 6,3 + 3,1
0,2x = 9,4
x = 9,4 : 0,2
x = 47
Th2:
0,2x - 3,1 = - 6,3
0,2x = - 6,3 + 3,1
0,2x = - 3,2
x = - 3,2 : 0,2
x = - 16
Vậy x = 47 hoặc x = - 16
b) |12,1x + 12,1 . 0,1| = 12,1
|12,1(x + 0,1)| = 12,1
\(12,1\left(x+0,1\right)=\pm12,1\)
Th1:
12,1(x + 0,1) = 12,1
x + 0,1 = 1
x = 1 - 0,1
x = 0,9
Th2:
12,1(x + 0,1) = - 12,1
x + 0,1 = - 1
x = - 1 - 0,1
x = - 1,1
Vậy x = 0,9 hoặc x = - 1,1
c) |0,2x - 3,1| + |0,2.x + 3,1| = 0
|0,2x - 3,1| + |0,2x + 3,1| \(\ge\) |0,2x - 3,1 + 0,2x + 3,1| = 0,4x
mà |0,2x - 3,1| + |0,2.x + 3,1| = 0
=> x = 0
x + x : 0,2 = 1,35
x * 1 + x * 5 = 1,35
x * ( 1 + 5 ) = 1,35
x * 6 = 1,35
x = 1,35 : 6
x = 0,225
hok tốt nha ^_^
| 12,1 .x + 12,1 . 0,1 | = 12,1
| 12,1 . ( x + 0,1 ) | = 12,1
\(\Rightarrow\)| x + 0,1 | = 12,1 :12,1
| x + 0,1 | = 1
| x | = 1 - 0,1
| x | = 0,9
x = 0,9
Vậy x = 0,9